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Math Help - marginal probability density function

  1. #1
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    Question marginal probability density function

    fX,Y(x,y) = 6(1-x-y) for x and y defined over the unit square, subject to the restriction that 0<x+y<1. Find the marginal probability function for X.

    I have started by integrating with limits 1-x and -x getting a final answer of 18(x-1-2x^2). Anyone else get this answer?

    Thanks
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by ben.mahoney@tesco.net View Post
    fX,Y(x,y) = 6(1-x-y) for x and y defined over the unit square, subject to the restriction that 0<x+y<1. Find the marginal probability function for X.

    I have started by integrating with limits 1-x and -x getting a final answer of 18(x-1-2x^2). Anyone else get this answer?

    Thanks
    Since you integrate over the unit square, x \in [0,1] and y \in [0,1]
    so the integral over y should go from 0 to 1-x.

    And this gives something like 3(1-x)^2
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    do you get a final answer of  18(x^3-3x^2+3x-1)<br />
    Last edited by mr fantastic; March 17th 2009 at 05:40 AM. Reason: Fixed latex tags
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  4. #4
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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    do you get a final answer of  18(x^3-3x^2+3x-1)<br />
    Is that the same as the answer moo gave you ....?

    Quote Originally Posted by Moo View Post
    Hello,

    Since you integrate over the unit square, x \in [0,1] and y \in [0,1]

    so the integral over y should go from 0 to 1-x.

    And this gives something like {\color{red}3(1-x)^2}
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    i get  -3(1-x^2) but you have to multiply this by the constant (6-6x) which gives my answer?
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  6. #6
    Moo
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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    i get  -3(1-x^2) but you have to multiply this by the constant (6-6x) which gives my answer?
    It would help a lot if you show your working. It'd be far easier to spot out your mistake.
    I've just made the integration again, and I still found 3(1-x)... :s
    If Mr F didn't say anything, then there is some chance that it is correct
    And why do you want to multiply by (6-6x) ? o.O
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  7. #7
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    integrate 6[1-x-y] from 0 to 1-x
    you can take the (6-6x) out the front of the integral as its just a constant.
    this gives (6-6x) integral -6y
    integrating -6y gives -3y^2
    putting 1-x in makes the -3(x-1)^2 multiplied by the constant (6-6x)
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  8. #8
    Moo
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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    integrate 6[1-x-y] from 0 to 1-x
    you can take the (6-6x) out the front of the integral as its just a constant.
    this gives (6-6x) integral -6y
    integrating -6y gives -3y^2
    putting 1-x in makes the -3(x-1)^2 multiplied by the constant (6-6x)
    (6-6x) is not a constant that you simply pull out from the integral... And it doesn't even multiply the integrand !

    If a is a constant, then \int af=a \int f, but that's all you can say !

    You can pull out 6, nothing else.
    \int_0^{1-x} 6(1-x-y) ~dy=6 \int_0^{1-x} (1-x)-y ~dy=I
    It is true that (1-x) is a constant, but only with respect to y.

    You can separate the integral :
    I=6 \int_0^{1-x} (1-x) ~dy-6 \int_0^{1-x} y ~dy
    I=6(1-x) \int_0^{1-x} 1 ~dy-6 \int_0^{1-x} y ~dy
    I=6(1-x) \times \left. y \right|_{y=0}^{y=1-x} -6 \times \left. \frac{y^2}{2} \right|_{y=0}^{y=1-x}

    I=6(1-x) [(1-x)-0]-6 \left[\frac{(1-x)^2}{2}-\frac 02\right]

    ...
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